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A Precise Theory for the Proton Radius and the Elements as Mathematical Sets!
Fourth Edition!
By!
Ian Beardsley!
Copyright © 2023"
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Contents!
Preface……………………………………………………….3!
Data…………………………………………………………..4!
Abstract………………………………………………………5!
Introduction………………………………………………….6!
1.0 Ground Work……………………………………………6!
2.0 The Elements As Proton Sets….. …………………..10!
3.0 Solving The Calendar………………………………….14!
4.0 Predicting The Meter…………………………………..23!
5.0 The Constant k…………………………………………26!
6.0 Discussion……………………………………………..35!
7.0 Solar System A Quantum Mechanical System…….36!
8.0 Solar Luminosity A QM State………………………..43!
9.0 Genesis Project Equations……………………………47!
10.0 Another Theory For The Radius Of A Proton……..49!
11.0 The Macro and Micro Planck Constants………….53!
12.0 The Exact Radius of a Proton………………………57!
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Preface
One of things I wanted to go into in this paper but didn’t is that you can speak of the structure of
the solar system even though it changes with time. This is important to understand when I refer
to sizes the Moon and the planets, and their orbital distances.
The whole object of developing a theory for the way planetary systems form is that they meet the
following criterion: They predict the Titius-Bode rule for the distribution of the planets; the
distribution gives the planetary orbital periods from Newton’s Universal Law of Gravitation. The
distribution of the planets is chiefly predicted by three factors: The inward forces of gravity from
the parent star, the outward pressure gradient from the stellar production of radiation, and the
outward inertial forces as a cloud collapses into a flat disc around the central star. These forces
separate the flat disc into rings, agglomerations of material, each ring from which a different
planet forms at its central distance from the star (it has a thickness). In a theory of planetary
formation from a primordial disc, it should predict the Titius-Bode rule for the distribution of
planets. When I speak of the state of the solar system I speak of this point toward which the
solar system has formed and not the small changes that happen over millions of years due to
mutual interference between the bodies. In fact, mutual interference has torn apart possible
forming planets resulting in the current distribution we have today, because the current
distribution is more or less stable. The asteroid belt is a good example of this — it is a location
where a planet cannot form due to harmonic (repetitive) action on the orbital period at its
distance by orbital periods of planets beyond it. In short we take the state of the solar system as
an inflection point between what it became, and what it might minutely be going away from in
billions of years after it dies and can no longer support life.
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The natural constants measure the properties of space and time. We can list some of them here:
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
m
P
: 1.67262 × 10
−27
kg
h : 6.62607 × 10
−34
J ⋅ s
r
p
: 0.833 × 10
−15
m
G:6.67408 × 10
−11
N
m
2
kg
2
α : 1/137
q
p
= q
e
= 1.6022E − 19coulom bs
k
e
= 8.988E 9
Nm
2
C
2
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In this the fourth edition in section 12.0 we suggest if the nature of space and time is such the
radius of a proton is given by the golden ratio, then we can determine its value exactly in theory
and we find this in turn gives the momentum-distance of quantum mechanics, and the
asymptote of the special relativity hyperbola of spacetime diagrams that is the world-line of a
light signal.
In the third edition we create in section 11.0 a Planck constant on the macro-scale and compare
it to the Planck constant of quantum mechanics and see that the Earth/Moon/Sun system is a
quantum mechanical system as well because the kinetic energy of the Earth is an eigenvalue
mapping the macro-scale Planck constant to the micro-scale Planck constant, as well as the
Planck constants being eigenvalues as in quantum mechanics for proton/electron systems but
here for the Earth kinetic energy around the Sun. This is done by setting out to show that the
second is a natural unit, a natural constant, which we achieve.
The proton is the fundamental charged particle that makes up atoms. We develop a theory of
inertia with the proton as a 3d cross-section of a 4d hypersphere and suggest in terms of set
theory that since no proton is unique that number of protons is actually number. This describes
the elements of the periodic table as sets of protons in a mathematical sense, and we suggest this
is geared around six-fold symmetry in a way that optimizes the conditions for biological life. The
radius of a proton is the least precise data value among the constants but new methods of
measuring it are honing in on its actual value. If our set theory for the elements in terms of
protons optimizes for life, then we have from our theory a value that is more within the current
finest measurements. We also suggest the unit of a second is a natural unit as luck would have it,
and that it occurs in the orbital elements of the Earth/Moon/Sun system. We determine a
constant k and intermediate mass that bridges the macrocosmos to the microcosmos that
predicts many things both on the atomic and astronomical and cosmological scales. It appears
our moon orbiting the earth may have connections on the atomic and cosmological scales. We
present the Solar System as quantum mechanical. In the second edition we come up with
another equation for the radius of a proton independent of Avogadro’s number which gives us
more equations for the unit of a second as natural to relate to the orbit of our Moon."
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Introduction
The old enigma, the Universe is here, but it needs a cause. If we present a cause, then when
need a cause for that cause. Essentially it is as if the Universe is magic because it should not
be here; it is an uncaused cause. But, can we see that that can happen in terms of
mathematics?!
I had a theory for what matter is, or inertia. Mass is that which fills space over time. What is
space and time? We don’t know, it is our starting point and is not described by anything. For
instance what is velocity? It is the change in distance with time, meters per second (m/s). Let
us write velocity:!
Equation 1. !
But what is the distance x? It is velocity over time, that is how we define it. Let’s write that:!
Equation 2. !
Let’s put our value of velocity into equation 2:!
!
We have t cancels with t leaving x=x meaning distance is distance, so we have not said
anything about distance (spacial extension in one of the dimensions). We can in fact put all of
the relationships between mass, length, and time in any equation in physics and it will return!
!
!
!
So in the end we may have formally written out for instance an equation for gravity that will
land us on the Moon, but the equation reduces to saying nothing about what reality actually is. !
Back to my theory that is my theory for inertia. That theory is another subject, but I would like
to show how it suggests we can have something from nothing, which makes me wonder if this
mathematical trick I have found, is connected to how we might actually have something from
nothing, or that perhaps even something is nothing. Let’s lay the groundwork of that theory
briefly:!
1.0. Ground Work
We suggest the property of matter to have inertia, which is to say it resists change that the
particle a proton, a fundamental unit of matter, is a the cross-section of a 4D hypersphere in 3D
space. As such one could consider inertia as the “friction” of space given by the normal vector
holding the space-bubble in a lower dimensional “plane”:!
Thus the mass of a proton is given by the force of space measured by G, the energy given
to it given by h, the speed at which things happen c, and the surface area of this sphere
v = x /t
x = vt
x =
x
t
t
m a ss = m a ss
distance = distance
time = time
m
p
4πr
2
p
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that is the cross-section of the hypersphere and the fine structure constant squared
because the fine structure constant squared is the ratio of the potential energy of an electron in
the first circular orbit to the energy given by the mass of an electron in the Bohr model times the
speed of light squared, that is it represents the ground state. It is
Equation 1.1
Thus the equation of inertia can be expressed as proton-seconds. We find that it is six-fold in
nature given the smallest integer for time (1-second) gives carbon the core element of life
chemistry, and 6-seconds is the largest integer before you get fractional protons. That is
Equation 1.2.
Equation 1.3.
From which instead of saying the left sides of these equations are seconds, we say they are
proton-seconds by not letting cancel with the bodies of these equations on the left, but
rather divide into them, which are in units of mass, giving us a number of protons. I say this is
the biological because as we shall see our equations are based on one second is 6 protons is
carbon, and 6 seconds is one proton is hydrogen, these making the hydrocarbons which are
the skeletons of biological life. We see this is a mystery of six-fold symmetry based around
biological life in the following computer program I wrote and its output:!
α
2
α
2
=
U
e
m
e
c
2
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton ⋅ 6secon d s = hydrogen(H )
m
p
1
α
2
m
p
h4π r
2
p
Gc
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Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by
the mass of a proton. But these masses can be considered to cancel and leave seconds. We
make a program that looks for close to whole number solutions so we can create a table of
values for problem solving.
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest
of the elements heavier than carbon are formed by fractional seconds. These are the
hydrocarbons the backbones of biological chemistry. Here is the code for the program:
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#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in equation 1 (no more than 100?): ");
scanf("%i", &n);
}
while (n>=101);
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);
}}}}
We find exactly our equation predicts the second as!
Equation 1.4 !
That this equals so perfectly one second leads us to suggest the second is a Natural Unit. We
find it is, that it is in the kinetic energy of the Moon to that of the Earth times the Earth Day
which is the Earth rotation period:!
Equation 1.5. !
But that was using the average orbital velocities. We find using aphelions and perihelions:!
Equation 1.6.
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
KE
moon
KE
Earth
(EarthDay) = 1.2secon d s
K E
moon
K E
earth
(Ear th Day) = 1.08second s
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2.0. The Elements As Proton Sets
To get the Universe from nothing we consider equation 1.2 and 1.4:!
Equation 1.2. !
Equation 1.4. !
We can consider equation 1.2 to be two equations. The first is where the numerator!
!
Cancels with the denominator!
!
Leaving:!
Equation 2.1. !
And the second equation is where does not cancel with the numerator but divides into it
giving us a number of protons times one second:!
Equation 2.2. !
Now we find if we divide the left hand side by one second ( ) then the right hand side becomes
six protons.!
Equation 2.3. !
Now consider equation 1.4!
1
α
2
m
p
h4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
1
α
2
h4π r
2
p
Gc
= k ilogra m s ⋅ secon d s
m
p
1
α
2
m
p
h4π r
2
p
Gc
= 6secon d s
m
p
1
α
2
m
p
h4π r
2
p
Gc
= 6proton − secon d s
t
1
1
t
1
1
α
2
m
p
h4π r
2
p
Gc
= 6proton s
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!
It is approximately . Let us put it into equation 2.3:!
Equation 2.4. !
!
However factor 1!
!
Cancels with factor 2!
!
When this happens it says 6 equals 6 protons:!
!
The connection between number and form and they are the carbon atom that is at the core of
life. We can do the same thing with hydrogen (Equation 1.3):!
We see carbon is given by one second and hydrogen is given by six seconds ( equation 1.2)
Thus they are out of phase with one another by a factor of 6. We see this six-fold symmetry
produced hydrogen and carbon from nothing, and that they may in fact exist, but be nothing,
that they make up the skeletons of life that are the hydrocarbons, used as the basic chemical
structure underlying our mind that is consciousness. Thus we are conscious, but don’t really
exist. We seem to be conscious in a universe where the atom, the proton, is described by one
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
t
1
= 1secon d
6
α
2
m
p
1
Gc
h4π r
2
p
1
α
2
m
p
h4π r
2
p
Gc
=
6proton ⋅ secon d s
1secon ds
= 6proton s
6
α
2
m
p
1
Gc
h4π r
2
p
1
α
2
m
p
h4π r
2
p
Gc
6 = 6protons
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton ⋅ 6secon d s = hydr ogen(H )
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
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second, which came historically from our way of dividing up the calendar in ancient times
without knowledge of the atom in terms of the these modern constants. We also see this second
exists not just with the atom and its proton but with the Earth/Moon/Sun system which I found
was equations 1.5 and 1.6:
Equation 1.5. !
Using the average orbital velocities. We find using aphelions and perihelions:!
Equation 1.6.
Earth day=(24)(60)(60)=86,400 seconds.
Using the Moon’s orbital velocity at aphelion, and Earth’s orbital velocity at perihelion we have:
Using the Moon’s orbital velocity at aphelion, and Earth’s orbital velocity at perihelion. We can
even use not the solar day used here (rotation of Earth with respect to the Sun) but the sidereal
day (rotation of Earth with respect to the stars) and the equation has even slightly more accurate
results.
In my work A Contemporary Anthropological Cosmology I derived the unit of a second in
terms of the hydrogen atom, and created a constant k that enabled me to predict the radius of a
proton with our equation 1.2 and 1.3 within experimental errors. But for now the point of this
paper is to show that either something can come from nothing, and that something is nothing,
and that this nothing actually creates life chemistry, if not physical chemistry in general, that
gives us consciousness, and that this consciousness has a connection with the Moon that it sees
in the sky from its planet Earth.
Interestingly the Moon perfectly eclipses the Sun which means as seen from the Earth the Moon
appears to be the same size as the Sun. This is because
Where is the Earth orbital distance, is the Moon’s orbital distance, is the Sun’s radius,
and is the Moon’s radius.
The Moon at its inclination to the Earth in its orbit makes life possible here because it holds the
Earth at its tilt to its orbit around the Sun allowing for the seasons so the Earth doesn’t get to
extremely hot or too extremely cold. We see the Moon may be there for as much of reason as is
Alpha Centauri, for a first manned mission beyond Earth.
KE
moon
KE
Earth
(Ear th Day) = 1.2secon d s
K E
moon
K E
earth
(Ear th Day) = 1.08secon d s
K E
moon
=
1
2
(7.347673E 22kg)(966m /s)
2
= 3.428E 28J
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
r
e
r
m
≈
R
⊙
R
m
r
e
r
m
R
⊙
R
m
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If every proton in the universe is the same, none having any distinguishable features, that is
none are unique because they all have the same mass, same size, and same charge, then six
protons is the number 6 is carbon, and 1 proton is the number 1 is hydrogen, and carbon being
the number 6 tells us what the number 6 is, and hydrogen being the number 1 tells us what the
number 1 is. Because quanta such as protons and electrons are so small that the Heisenberg
uncertainty principle applies, then our proton, for instance, has no position in space only a
probability of existing in a given region at a given time. If then the proton has no position, it
cannot be distinguished from another proton by where it is in space. Thus I suggest the chemical
elements are sets, where each proton is the element of a set. Carbon is the set of six protons, and
hydrogen is the set of one proton. Because sets by set theory cannot have repeating elements, we
cannot say the set of protons that is carbon is
But rather we have to say
And the set of protons that is hydrogen is
We need not put the identifier 1 onto proton because hydrogen can never have more than one
proton because if it does it ceases to be hydrogen and is another element, like helium if you add
another proton. Hydrogen is thus the number 1 and does not need an identifier and 6 (carbon)
does need identifiers. I find this interesting because hydrogen cannot be identified as a metal or
non-metal but carbon can be identified as a non-metal. When a non-metal like carbon combines
with hydrogen it loses an electron to become thus ionizing like a metal and carbon gains 4
electrons to have noble gas electron configuration the same as neon (Ne) the noble gas in group
18 the last group of its period. But since carbon gains 4 electrons to become it must
combine with at least 4 hydrogens each to be neutral. This is which is methane gas.
This is C+4H=10 protons which is the element neon (Ne). This gives a time value of
Methane is the most basic hydrocarbon and hydrocarbons are the skeleton of biological life
chemistry. Methane gas was in the primordial Earth and when combined with ammonia ,
, and other gases thought to have been in the primordial Earth atmosphere, they make
some of the amino acids, the building blocks of life as was shown in the Miller-Urey experiment.
The radius of a proton is the least well known constant in our equation and is the subject of
much research to determine precisely. If our value of time t is supposed to be precisely
So as to center for it allowing for biological life then the radius of a proton is
C =
{
proton, pr oton, proton, pr oton, proton, pr oton
}
C =
{
proton1,pr oton2,proton3,pr oton 4,pr oton5,pr oton6
}
H =
{
proton
}
H
1+
C
4−
H
1+
CH
4
t =
1
10
⋅
187769
1.67262E − 27
(6.62607E − 34)
6.67408E − 11
⋅
(0.833E − 15)
2
(299792459
= 0.602997811s
NH
3
H
2
O
H
2
0.6 =
6
10
=
carbon
neon
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This is in the range of the PRad (JLab) electron scattering method, a most recent method for
determining the radius of a proton, which is
3.0 Solving The Calendar Thus the duration of a second is!
!
Thus we developed a system to measure time from the motion of the Earth around the
Sun that creates a system where we measure time with a second that is 3 hundred
millionths of a year, the time it takes the Earth to go around the Sun once, and where
there are 86400 of these seconds (solar day) in the time it takes the Earth to rotate
once on its axis with respect to the Sun. With respect to the stars it takes 86164.0905
seconds. This is called the sidereal day. The solar day is a little longer because the
Earth has moved around the Sun a little before the Earth aligns with it again in its
rotation. It is this duration of a second that has given us!
Where 6 seconds is the largest integer before we get fractional atoms and 1 second is the
smallest integer that produces atoms and hydrogen and carbon are the hydrocarbons that make
the chemical skeletons of life chemistry.
!
That this equals so perfectly one second leads us to suggest the second is a Natural Unit. We
find it is, that it is in the kinetic energy of the Moon to that of the Earth times the Earth Day
which is the Earth rotation period:!
!
But that was using the average orbital velocities. We find using aphelions and perihelions:!
r
p
=
(6.67408E − 11)(299792459)
(6.62607E − 34)4π
10
18769
⋅ (1.67262E − 27 )(0.6) = 8.2885873E − 16m ≈ 0.829E − 15m
r
p
= 0.831E − 15m
±
2 %
1second =
1
365.25
⋅
1
24
⋅
1
60
⋅
1
60
= 0.0000000316881year
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton ⋅ 6secon d s = hydr ogen(H )
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
KE
moon
KE
Earth
(Ear th Day) = 1.2secon d s
of 15 61
These equations derive their meaning from classical and modern physics, yet the second came to
us from Ancient history of lost civilizations. We wish to present that as a mystery to add on to
these discoveries about Nature. We begin
It takes the Moon 27.3 days to go around the Earth, but since the Earth has orbited the Sun
since then it takes 29.5 days with respect to the Sun, or from New Moon to New Moon in other
words. Thus we say the lunar month is 30 days, the average of the 29 day month and the 31 day
month. I think I have solved the Calendar and why it is structured the way it is, why there are
365.25 days in a year. Well we know why that is, the earth rotates 365.25 times after it has gone
around the Sun once. But I think when making the calendar from which we got the second we
considered that 365.25 days is approximately 360 days, because we divide the circle into 360
degrees.
At the dawn of civilization, the Sumerians who settled down from wandering, gathering, and
hunting, to invent agriculture and and build ceramic homes, developed the first mathematics,
and it was sexagesimal (base 60) that passed on to the Babylonians, then ended up with the
Ancient Greeks, who divided not just the hour into 60 minutes of time and the minute into 60
seconds of time, but who divided the sky into hours, minutes, and seconds of arc which was
measured in time counted by the rotation of the Earth, and its orbital period around the Sun,
and the orbital period of the Moon around the Earth.
We are suggesting nature is founded on six-fold symmetry, six which is constructed by
multiplying together the first two prime numbers, 2 and 3. Two factorial is two, three factorial is
6. Two times six is twelve, the number of months in a year, approximated by our moon’s
approximately 12 orbits around the earth in the time it takes the earth to go around the Sun
once. There are four weeks in a lunar month, and
Is the number of seconds in the 24 hour Earth day. The duration of one second comes from
dividing up time and the sky like this in base 60, that came from the Sumerians, Babylonians,
and Ancient Greeks and it is believed they did this because 60 is evenly divisible by
1,2,3,4,5,6,…10,12,15,20,30,60,…
I think I have solved the calendar and why it is structured the way it is, and that it is not just
sexagesimal where 60 seconds and 60 minutes are concerned, but all the way through. We have
found the second pivotal to the atom and the Earth/Moon/Sun System in terms of sciences that
came long after the calendar’s formation. But, if we want to suggest the calendar was influenced
by advanced life from another planet, or that the Ancients lucked out because there was an idea
behind it, and we want to connect it to our equations of the proton and of the Moon and Earth
kinetic energies, which are dynamic, we have to find out why it is dynamic. I begin with the
second in terms of the calendar:
We write:
K E
moon
K E
earth
(Ear th Day) = 1.08secon d s
1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 60
2
= 86,400
1secon d =
1
365.25
⋅
1
24
⋅
1
60
⋅
1
60
= 0.0000000316881year
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There are actually 31557600 seconds per year, but we approximated 365.25 days with 360 days.
This however has an accuracy of
I then factor this into other ratios but keep the units:
Earth rotates through:
Moon moves around Earth:
Earth moves around Sun:
Thus our equation becomes
We then notice in one hour the earth rotates through 15 degrees giving
It does the same in 60 minutes. We also have
Seconds in a day: , and we write
(
360d a ys
year
)(
24h ours
d a y
)(
60min
h our
)(
60sec
min
)
= 31104000secon d s /year
31104000
31557600
= 98.56
12
2
(
60d a ys
year
)(
1h our
d a y
)(
60min
h our
)(
60sec
min
)
1h our = 60min
60d a ys =
1
6
year
360
24
=
15
∘
h our
360
∘
30d a ys
=
12
∘
d a y
360
∘
360d a ys
=
1
∘
d a y
60d a ys = 2(30d ays) = Lu n arOrbit
(
12d eg
d a y
⋅
d a y
deg
)
2
(
60d a ys
year
)(
1h our
d a y
)(
60min
h our
)(
60sec
min
)
24h ours
360deg
⋅ 15deg = 1h our
2(60
∘
) = 120
∘
2(120
∘
) = 240
∘
4(60
∘
) = 240
∘
(24h ours)(60min)(60sec) = 86400
sec
d a y
of 17 61
We have
1
86400sec
⋅ 360
∘
=
1
240
degrees
(
12d eg
d a y
⋅
d a y
deg
)
2
(
60d a ys
year
)(
24h ours
360
∘
⋅ 15
∘
)(
60min
h our
)(
60sec
min
)
(
60min
h our
⋅
60sec
min
)
=
3600sec
h our
1
3600
⋅ 360 =
1
10
degrees
(
24h ours ⋅
15deg
360deg
)(
60min
h our
)(
60sec
min
)
= 86400sec ⋅
15
∘
360
∘
15/360 = 0.013889
1/x = 72
86400
72
= 1200secon d s
of 18 61
Thus we have reduced the second to special triangles. The same second in our equation for the
proton, and for the kinetic energies of the Earth and Moon, and the rotation of the Earth. If we
can explain all three of these; the proton, the kinetic energies, and structural evolution of the
calendar, in terms of these triangles, we can speak of how they are all part of an underling idea
common to all three.
We have the following expression for seconds in a year:
We write out the angular velocities of the Moon and Earth in their orbits:
Put in the orbital periods of the Moon and the Earth and :
(
12d eg
d a y
⋅
d a y
deg
)
2
(
60d a ys
year
)(
24h ours
360
∘
⋅ 15
∘
)(
60min
h our
)(
60sec
min
)
·
θ
m
=
12d eg
d a y
·
θ
e
=
deg
d a y
(
·
θ
m
·
θ
e
)
2
(
2 ⋅ 30d a ys
year
)(
24h ours
360
∘
⋅ 15
∘
)(
60min
h our
)(
60sec
min
)
T
m
T
e
(
·
θ
m
·
θ
e
)
2
(
2
T
m
T
e
)
(
T
e
⋅
15
∘
360
∘
)(
60min
h our
)(
60sec
min
)
(
·
θ
m
·
θ
e
)
2
(
T
m
T
e
)
(
T
e
⋅
30
∘
360
∘
)(
60min
h our
)(
60sec
min
)
(
·
θ
m
·
θ
e
)
2
(
T
m
)
(
30
∘
6
∘
)(
1min
h our
)(
60sec
min
)
(
·
θ
m
·
θ
e
)
2
(
T
m
) (
5
)
(
60sec
h our
)
(
·
θ
m
·
θ
e
)
2
(
T
m
)
(
300sec
h our
)
1h our = 3600secon d s
of 19 61
=
Which is true because it is
And we had made the approximation that there are 360 days in year because there are 360
degrees in a circle. We see the calendar divides units of measurement into base 60
Which describes the motion of the moon and the Earth
In the literature it says we use base 10 probably because we have 10 fingers to count on. But 10 is
2 times 5 and 2 is the smallest prime number. Base 60 is 10 times six and sixfold symmetry is
the most dynamic because it is 2 times 3 and 2 and 3 are the smallest prime numbers one even,
one odd. The regular hexagon, a six-sided polygon has its radii equal to its sides because it
divides-up into equilateral triangles. But I suggest we have base 10 not just because we have 10
fingers to count on but because 2 times five is 10 and 10 times 6 is 60 and 60 is evenly divisible
by so much. Evenly divisible by
1,2,3.4.5.6….10,12,15,20,30,60
It is known that the physical like snowflakes are based on 6-fold symmetry and life is based on
5-fold symmetry like starfish or two arms, two legs, and a head. More often than not a flower
will have 5 petals. If it has hundreds like a rose, it will be based on the golden spiral and the
golden ratio is derived from the regular pentagon (five-sided polygon). Thus we have not just
solved the calendar, but found more reasons why we have base 10 counting. It alls has to do the
physical, that of six, and the organic, that of five.
(
·
θ
m
·
θ
e
)
2
(
T
m
)
(
300sec
3600sec
)
(
·
θ
m
·
θ
e
)
2
(
T
m
)
(
1
12
)
·
θ
m
·
θ
e
= 12
(
·
θ
m
·
θ
e
)
(
T
m
)
=
secon d s
year
(
12
)
(
30d a ys
)
= 360d a ys
(
12d eg
d a y
⋅
d a y
deg
)
2
(
60d a ys
year
)(
24h ours
360
∘
⋅ 15
∘
)(
60min
h our
)(
60sec
min
)
(
·
θ
m
·
θ
e
)
(
T
m
)
=
secon d s
year
of 20 61
Since we have
Then it follows
We want to write this in terms of the sizes of the Moon and the Sun, the masses of the Earth and
the Sun, and the orbital radii of the Moon and the Earth.
Essentially that the Moon perfectly eclipses the Sun as seen from the Earth means while it is 400
times smaller than the Sun it is 400 times further from the Sun than it is from the Earth. This
determines its orbital velocity and mass, as well as that of the Earth and the mass of the Sun.
The orbital velocities of the Moon and the Earth are given by:
and
The Moon perfectly eclipses the Sun because
Where is the Earth orbital radius, is the lunar orbital radius, is the solar radius, and
is the lunar radius. This gives:
We have more explicitly:
K E
moon
K E
earth
(Ear th Day) = 1.08secon d s
(
·
θ
m
·
θ
e
)
(
T
m
)
=
secon d s
year
(
·
θ
e
·
θ
m
)
(
T
e
T
m
)
(
K E
m
Ke
e
)
(
Ear th Day
)
= 1secon d
v
e
=
GM
⊙
r
e
v
m
=
GM
e
r
m
v
e
v
m
=
M
⊙
M
e
⋅
r
m
r
e
M
⊙
M
e
⋅
r
m
r
e
=
1.989E 30kg
5.972E 24
⋅
1.74E6m
6.957E8m
= 28.86
r
e
r
m
≈
R
⊙
R
m
r
e
r
m
R
⊙
R
m
v
e
v
m
=
M
⊙
M
e
⋅
R
m
R
⊙
= 28.6
of 21 61
The rotational period of the moon is is the same as its
orbital period, and the rotation of the Earth is is the ratio
27.3 which gives
From
We have
We want to see how accurate this is:
But the radius of the Sun is difficult to pinpoint. Beyond the bulk of its mass it has a thin
atmosphere that reaches far out into space. However, the Moon nearly perfectly eclipses the Sun
as seen from the Earth allowing us to study its outer atmosphere during an eclipse. The moon
nearly perfectly eclipses the Sun as seen from the Earth because the Sun may be 400 times
larger than the Moon, but the Moon is 400 times further from the Sun than it is from the Earth.
This is to say that
Where is the Earth orbital radius, is the lunar orbital radius, is the solar radius, and
is the lunar radius. So why not let the size of the Moon as seen from the Earth define the radius
of the Sun by writing
v
e
v
m
=
R
m
R
⊙
M
⊙
M
e
⋅
r
e
r
m
R
m
R
⊙
M
⊙
M
e
⋅
r
e
r
m
=
1.74E6m
6.957E8
1.989E 30kg
5.972E 24kg
⋅
1.496E11m
3.84748E8m
= 28.46
T
m
= 27.3d ays = 2,358, 720secon d s
Γ
e
= 24hours = 86,400secon d s
T
m
Γ
e
=
R
m
R
⊙
M
⊙
M
e
⋅
r
e
r
m
(
·
θ
e
·
θ
m
)
(
T
e
T
m
)
(
K E
m
Ke
e
)
(
Ear th Day
)
= 1secon d
T
e
Γ
e
⋅
(
·
θ
e
·
θ
m
)
R
⊙
R
m
M
e
M
⊙
⋅
r
m
r
e
⋅
K E
m
K E
e
(
Ear th Day
)
= 1secon d
31557600s
86400s
⋅
(
·
1
·
12
)
7.1544E8m
1.7371E6m
5.972E 24kg
1.989E 30kg
⋅
3.6323E8m
1.496E11m
(1.08s) = (1.07)(1.08s) = 1.1556 s
r
e
r
m
≈
R
⊙
R
m
r
e
r
m
R
⊙
R
m
R
⊙
=
r
e
r
m
R
m
of 22 61
The Moon at closest approach to Earth (perigee) is 1.7371E7 meters:
This is the radius of the Moon I used to verify our equation. But let us take
And verify it with
Equal to its exact value for the Earth year, and the sidereal month of the moon.
=0.9856deg/day
And our results just become better. Now we just have one more thing to do, to take our equation
for the second in terms of the proton
!
And!
To get,
1.496E11m
3.632289E8m
⋅ 1.7371E6m = 7.154446E8m
T
e
Γ
e
⋅
(
·
θ
e
·
θ
m
)
R
⊙
R
m
M
e
M
⊙
⋅
r
m
r
e
⋅
K E
m
K E
e
(
Ear th Day
)
= 1secon d
⋅
(
·
θ
e
·
θ
m
)
θ
e
=
1
365.25d a ys
360
∘
θ
m
=
1
27.3d a ys
360
∘
= 13.1868deg /d a y
31557600s
86400s
⋅
(
·
1
·
13
)
7.1544E8m
1.7371E6m
5.972E 24kg
1.989E 30kg
⋅
3.6323E8m
1.496E11m
(1.08s) = (0.988)(1.08s) = 1.067s
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
T
e
Γ
e
⋅
(
·
θ
e
·
θ
m
)
R
⊙
R
m
M
e
M
⊙
⋅
r
m
r
e
⋅
K E
m
K E
e
(
Ear th Day
)
= 1secon d
T
e
Γ
e
⋅
(
·
θ
e
·
θ
m
)
R
⊙
R
m
M
e
M
⊙
⋅
r
m
r
e
⋅
K E
m
K E
e
(
Ear th Day
)
=
1
6α
2
⋅
r
p
m
p
h 4π
Gc
of 23 61
4.0 Predicting The Meter In the last section we took our equation for a second as given by
the calendar!
And found we could write it in sexagesimal (base 60)
Which resulted in
Which we found was
Because
We then used that the Moon perfectly eclipses the Sun
And Universal gravity for the Earth and Moon
and.
To get
!
And since we had for the proton!
!
1secon d =
1
365.25
⋅
1
24
⋅
1
60
⋅
1
60
= 0.0000000316881year
12
2
(
60d a ys
year
)(
1h our
d a y
)(
60min
h our
)(
60sec
min
)
(
·
θ
m
·
θ
e
)
2
(
2 ⋅ 30d a ys
year
)(
24h ours
360
∘
⋅ 15
∘
)(
60min
h our
)(
60sec
min
)
(
·
θ
m
·
θ
e
)
(
T
m
)
=
secon d s
year
(
12
)
(
30d a ys
)
= 360d a ys
r
e
r
m
≈
R
⊙
R
m
v
e
=
GM
⊙
r
e
v
m
=
GM
e
r
m
T
e
Γ
e
⋅
(
·
θ
e
·
θ
m
)
R
⊙
R
m
M
e
M
⊙
⋅
r
m
r
e
⋅
K E
m
K E
e
(
Ear th Day
)
= 1secon d
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
of 24 61
We ended-up with!
We now want to put certain quantities in parenthesis:
We have the key orbital elements and the equation of a proton:
We want to consider
411.858 refers to the Sun being 400 times larger than the Moon allowing for an eclipse because
the Moon is 400 times further from the Sun than from the Earth. Since we have rounded the
year to 360 days because there are 360 degrees in a circle and we have rounded the lunar
angular orbital velocity to 12 degrees and that of the Earth to 1 degree, both per day, then let us
write
Then 0.000085383 becomes x=0.000083333 and we have
We notice that the radius of a proton is . Thus we have
T
e
Γ
e
⋅
(
·
θ
e
·
θ
m
)
R
⊙
R
m
M
e
M
⊙
⋅
r
m
r
e
⋅
K E
m
K E
e
(
Ear th Day
)
=
1
6α
2
⋅
r
p
m
p
h 4π
Gc
(
T
e
Γ
e
)(
·
θ
e
·
θ
m
)
R
⊙
R
m
M
e
M
⊙
⋅
r
m
r
e
(
K E
m
K E
e
(Ear th Day)
)
=
1
6α
2
⋅
r
p
m
p
h 4π
Gc
(
T
e
Γ
e
)
=
360d a ys
1d a y
(
K E
m
K E
e
(Ear th Day)
)
= 1secon d
(
·
θ
e
·
θ
m
)
=
1
∘
/d ay
12
∘
/d ay
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 1secon d
R
⊙
R
m
M
e
M
⊙
⋅
r
m
r
e
=
7.15448m
1.7371E6m
5.972E 24kg
1.989E 30kg
⋅
3.6323E8m
1.496E11m
= (411.8588)(0.000085382)
(
360
)
(
1
12
)
(
400
) (
x
) (
1sec
)
= 1sec
M
e
M
⊙
⋅
r
m
r
e
= 0.000083333
r
p
= 0.833E − 15m
0.000083333
r
p
= 1.000E11m
−1
of 25 61
We know the average orbital distance of the Earth from the Sun is an astronomical unit (AU)
which is 1 AU=1.496E11m. That is because the orbit is close to circular. Thus we have
Using the orbital velocity of the Earth at perihelion (closest approach to the Sun when it is going
fastest) and the average orbital velocity of the Moon, , we have
As you can see
And you will find there is little difference of you use the sidereal lunar month which is 86,164.1
seconds. You get 1.20687 seconds, which is even closer, in fact almost exact.
We call upon
!
And have!
r
e
= 1AU
M
⊙
M
e
⋅
r
e
r
m
r
p
r
e
=
1.989E 30kg
5.972E 24kg
⋅
1.496E11m
3.6323E8m
⋅ (0.833E − 15m)(1.496E11m) = 1.4595m
2
1.45954m
2
= 1.208m ≈ 1.2m
v
e
= 30,290m /s
v
m
= 1,022m /s
K E
m
K E
e
(Ear th Day) = 1.2s
K E
m
=
1
2
(7.3476731kg)(1,022m /s)
2
= 3.83726E 28J
K E
e
=
1
2
(5.972E 24kg)(30,290)
2
= 2.7396E 33J
3.83726E 28
2.7396E 33
(86,400) = 1.210s
(
M
⊙
M
e
⋅
r
e
r
m
⋅ r
p
r
e
)
1
2
KE
m
KE
e
(Ear th Day)
=
1.2m eters
1.2secon d s
= 1.0
m
s
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
(
M
⊙
M
e
⋅
r
e
r
m
⋅ r
p
r
e
)
1
2
KE
m
KE
e
(Ear th Day)
⋅
1
6α
2
r
p
m
p
h 4π
Gc
= 1.0m eters
of 26 61
5.0 The Constant k We introduce a constant k, which is determined by the velocity of the
Earth:!
5.1 !
Six for six-fold symmetry. To derive this we introduce Giordano’s relationship. !
5.2 !
The number on the left is the number on the right, but with different units. I found I could
eliminate the and gain six-fold symmetry by creating an equation of state for the periodic
table of the elements using Avogadro’s Number ( ). We say that!
5.3
And,
5.4
Which is basically true, hydrogen is one proton and the mass of the electron is nominal. For
every 6E23 protons of hydrogen, there is one gram. Then we always have
5.5
We can say for any element
Where is the number of protons in the element, so for carbon
Thus by equation 5.2 we have
5.6
k v
e
= 6
h(1 + α) ⋅ 10
23
= G
10
23
6.02E 23
N
A
= 6E 23
proton s
gr a m
ℍ = 1
gr a m
proton
N
A
⋅ 𝔼 = 6E 23
𝔼
N
A
=
Z ⋅ 6E 23proton s
Z ⋅ gr a m s
𝔼 =
Z ⋅ gr a m s
Z ⋅ proton s
Z
ℂ
ℂ =
6gr a m s
6proton s
N
A
=
6(6E 23proton s)
6gr a m s
N
A
ℂ = 6E 23
h
(1 + α)
G
⋅ N
A
𝔼 = 6.00kg
2
⋅
s
m
of 27 61
And we have introduced our 6 of six-fold symmetry. We need an intermediary mass between the
microcosmos the proton mass and the mass of a star M. We use as the limiting factor for the
mass of a star as the Chandrasekhar limit, the upper limit of mass for a white dwarf star to form
without collapsing into a blackhole star. It balances with its gravity by radiation pressure alone.
It is:
5.7
We say the intermediary mass, , is
5.8
Inserting 5.7 into 5.8, we have
5.9
Where we made the approximation . This results in
5.10
We can hone this by reintroducing 0.77 for 3/4
Thus precisely:
We have
We have honing our
Multiplying 6 by we have our value for k.
m
p
M ≤ 0.77
c
3
ℏ
3
G
3
N
m
4
p
= 1.41 ⊙
m
i
m
i
= Mm
p
m
i
=
3
2
c
3
ℏ
3
G
3
m
2
p
1/2
0.77 ≈ 3/4
m
i
=
3
2
c
3
h
3
8π
3
G
3
m
2
p
1/2
= 67.9943kg
3
2
= 0.8660
0.77 = 0.8775
0.8775
0.8660
= 1.01328
m
i
m
i
= (67.9943)(1.01328) = 68.897kg ≈ 69kg ≈ 70kg
1/m
2
i
of 28 61
5.11
This gives a value of k as
5.12
5.13
Putting this in equation 5.1 we have
5.14
Which is very close to our prediction of six-fold symmetry. is the average orbital
velocity of the Earth. I find this constant k predicts the duration of the Universe in the
standard Friedman Model. We can write
5.15
This is
5.16
The constants suggest a basic Universal Energy by dimensional analysis
5.17
So we can write
5.18
1 Earth year = (365.25)(24)(60)(60)=31557600 seconds
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼 =
1
773.5
⋅
s
m
k =
1
(68.897kg)
2
(6.62607E − 34) ⋅
(1.007299)
6.67408E − 11
⋅ 6.02E 23 = 0.001268291s /m
1
k
= 788.4626
m
s
k v
e
=
29790m /s
788.46m /s
= 6.145748
v
e
KE = m
i
(
1
k
)
2
KE = (68.897kg)(788.4626m /s)
2
= 42,831,358Joules
h
G
⋅
c
3
m
p
= 1.599298E 29J
6.62607E − 34
6.67408E − 11
⋅
299792459
3
1.67262E − 27
= 1.599298E 29J
h
G
⋅
c
3
m
p
m
i
(
1
k
)
2
(1secon d ) =
1.599298E 29J
42831358J
= 3.734E 21secon d s
of 29 61
The universe is theorized by standard models to die in 100 trillion years, which is when the last
stars born will die out. This is exactly 1E14 years. We have
5.19
!
We find as well that k predicts the Earth year. We suggest for some mass , we have
It is given by
5.20
For all practical purposes this is the mass of the Moon, which is exactly 7.34767E22kg.
We have:
3.734E 21s
31557600s
= 1.1832332E14years ≈ 1E14years
h
G
⋅
c
3
m
p
m
i
(
1
k
)
2
(1secon d ) = Li feSpa nUniverse
K E
Earth
K E
moon
(1secon d ) = E ar th Da y
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
M
h
G
⋅
c
3
m
p
M
(
1
k
)
2
(1secon d ) = 1E ar thYear
M =
h
G
⋅
c
3
m
p
(1Ear thYear)
(
1
k
)
2
(1secon d )
h
G
⋅
c
3
m
p
=
6.62607E − 27
6.67408E − 11
⋅
(299792459)
3
1.67262E − 27
= 1.599E 36J ≈ 1.6E 36J
1
k
2
= (788.4626)
2
= 621,673.27
1Ear thYear = (365.25)(24)(60)(60) = 31557600s
M =
1.599E 36
621673
⋅
1
31557600
= 8.15E 22kg
8.15E 22kg
7/34767E 22kg
= 1.10919516 ≈ 1.12m oon s
of 30 61
5.21
!
We also find that k predicts the radius of a proton. We have the radius of a proton is given by
carbon by evaluating at one second:
5.22
But to get that we have to multiply by one second and we need one second in terms of the atom
for a theory of the proton. I find we can do that…
5.23
5.24
Substitute for to get
5.25
Where is the Van Der Waals radius for a hydrogen atom. We have now introduced the
radius of a hydrogen atom . Our formulation of inertia as proton seconds is a
form of impulse. To change that to momentum we have to divide by a second. This radius of the
hydrogen atom is the Van Der Waals radius, which is the closest distance between two hydrogen
atoms non-covalently bound. It is 120 pm. Divide that by where 1/k is our constant
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
(1secon d ) = 1Ear thYear
KE
Earth
KE
moon
(1secon d ) = Ear th Day
1
6α
2
m
p
h4πr
2
p
Gc
= 1.004996352seconds
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
t
1
=
1
6α
2
⋅
h 4π r
2
p
Gc
t
1
=
1
6α
2
⋅
h 4π r
2
p
Gc
t
6
=
r
p
α
2
m
p
h 4π
Gc
R
H
/2
r
p
t =
R
H
2α
2
m
p
h 4π
Gc
R
H
R
H
= 1.2E − 10m
ck
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
1 + α
⋅
1
N
A
𝔼
≈
1
k
=
3
4
⋅
1
m
p
c
3
h
8π
3
G
⋅
1
N
A
𝔼
of 31 61
And we find
5.26
5.27
We have our equation for the radius of a proton
We only need to multiply it by to have the right units, and we get
5.28
Then suggest we picked up 9/8 in approximations which is close to one anyway so we write
5.29
We form constants:
And we have the Equation:
5.30
We can say that Avogadro’s number is not an arbitrary number because it is such that there are
twelve grams of carbon and carbon is 6 neutrons plus 6 protons equals 12. We now want to add
to this section the equation of the mass of a proton.
t
ck
=
3 2
16
⋅
6.626E − 34
6.674E − 11
⋅
18769
(1.6726E − 27)
2
⋅
1.2E − 10
6.02E 23
⋅
1
π
= 1.12secon d s
3 2
16
⋅
h
Gπ
⋅
1
α
2
m
2
p
⋅
R
H
N
A
𝔼
= 1.12secon d s
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
t /ck = 1secon d
r
p
=
9
8
2 ⋅
1
m
p
hc
4π
3
G
⋅
R
H
N
A
𝔼
r
p
=
1
m
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
r
p
= 8.26935E − 16m ≈ 0.827f m
r
p
=
1
kg
(
kg
m
2
s
)
(
m
s
)
(
kg
s
2
m
3
)
R
H
N
A
ℍ
=
1
kg
(
kg
)
m
k =
hc
2π
3
G
= 6.93E − 9kg
R
H
N
A
𝔼
= 1.99E − 34m
r
p
m
p
= k
R
H
N
A
𝔼
of 32 61
5.31
If our equation is right and we put it into natural units then the product should be close to
one:
Let us start with the units with which we are working:
And convert these to proton-masses and proton-radii:
Now we find k in these units:
Thus we have from equation 30:
!
m
p
=
1
r
p
hc
2π
3
G
⋅
R
H
N
A
𝔼
r
p
m
p
G =
m
3
kg ⋅ s
2
h = kg ⋅
m
2
s
c = m /s
G = 6.67408E − 11
m
3
kg ⋅ s
2
⋅ 1.67262E − 27kg(0.833E − 15m)
3
⋅ s = 193,131, 756
h = 6.62607E − 34kg ⋅
m
2
s
⋅
s
(0.833E − 15)
2
(1.67262E − 27kg)
= 5.71E 23
c =
(299,792, 459m /s)(1sec)
(0.833E − 15m)
= 3.6E 23
R
H
=
1.2E − 10m
0.833E − 15m
= 144,058
k =
hc
2π
3
G
= 6.93E − 9kg
k =
(5.71E 23)(3.6E 23)
2π
3
(193131756)
= 4E18proton − m a sses
r
p
m
p
= k
R
H
N
A
𝔼
r
p
m
p
=
(4E18)(144058)
(6E 23)
=
5.76E 23
(6E 23)
= 0.96 ≈ 1
of 33 61
We can also predict the charge of an electron with k…
I construct the electric field as such (Fig. 2): is one component of . Events in
are through time with components and where c is the speed of light. The
electrons and the protons and in the presence of one another cross into giving
themselves acceleration in the and directions.
We have
We suggest
5.32
The position of changes in the space of and has travelled to a place in
time, as well. Since ct=meters there is a g such that . Thus since
x
ℝ
3
(x, y, z)
ℝ
3
(x, y, z)
ct
1
ct
2
q
e
q
p
ct
1
ct
2
x
−x
i
j
k
0 ct
1
0
0 0 ct
2
= (ct
1
− ct
2
)
i
··
x ∝ (ct
1
− ct
2
)
i
x
ℝ
3
(x, y, z)
t (0,ct
1
, ct
2
)
··
x = g(ct
1
− ct
2
)
i
g ⋅ m =
m
s
2
Fig. 2
of 34 61
and that is . Thus g is frequency squared ( ) and we suggest it is
derived from the separation between the charges and . That there is some velocity v such
that . We call upon our equation for 1 second:
Letting v= from our equation for k
And we have
5.33
We get
5.34 !
g =
1
s
2
=
1
t
2
1
s
2
⋅
m
s
⋅ s =
m
s
2
f
2
q
e
q
p
f =
(
v
x
)
r
p
m
p
h 4π
Gc
= 1secon d
(
v
x
)
2
(c)
r
p
m
p
h 4π
Gc
=
k
e
m
⋅
q
2
x
2
v
2
(c)
r
p
m
p
h 4π
Gc
= k
e
q
2
m
α
2
6
⋅
1
k
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼 =
1
773.5
s
m
1
k
= 773.5
m
s
m = m
p
c
k
e
(
α
2
6
⋅
1
k
)
2
h 4π r
2
p
Gc
= q
2
h 4π r
2
p
Gc
=
(6.626E − 34)(4π)(0.833E − 15)
2
(6.674E − 11)(299,792, 459)
= 5.37E − 31
q = (0.033)
1
36(18769
2
)
773.5
2
(5.37E − 31) = 9.1435E − 19C
q
q
p
= ⋅
9.1435E − 19C
1.602E − 19C
= 5.71proton s ≈ 6protons
of 35 61
6.0 Discussion
When factor 1 and factor 2 cancels (page 11) it says 6 equals 6 protons. The connection between
number and form.
Interestingly this harkens back to the Logical Positivists. At the time mathematicians realized
nothing had any meaning that we didn't even know what 1, or 2, or 3, or 4....meant. So Bertrand
Russel set out to make a mathematics that explained it. It was said 4 was 4 objects and and set
theory was invented to show that a set was a collection of objects, for which its inventor Georg
Cantor made rules to explain what an object was. But in the end the British mathematician
Bertrand Russel showed it to contradict itself, then Gödel came along and proved that all formal
systems have at least one unprovable statement, and are thereby incomplete with his
Incompleteness Theorem. So they gave up. I think this is connected to 6 equals 6 protons in my
theory. They couldn’t say what an object was though it was tried to establish that it could be
anything but here it may be a proton and I think that works because any object is made of
protons, and protons are the basic units of matter and are uniformly the same. I am getting
interested now in this set theory and Gödel's theorem. But the recently added chapter on solving
the calendar has got me very intrigued as well, just where the Ancients got it, or if it evolved to
something naturally on its own.
The logical positivists may have given up on logical positivism, but set theory is used today in
applications even though it never said what an object is or a number. I think it is interesting that
I get 6 is 6 protons, is carbon or hydrogen is 1 is 1 proton because objects are made up of
protons. Interesting are the quarks which are inferred from experiment, but cannot owning to
the principle of color confinement, exist in isolation, they must be found in hadrons. Hadrons
are protons and neutrons. But the proton can exist in isolation.
of 36 61
7.0 Solar System A Quantum Mechanical System The constant k in this paper is a sort of
intermediate velocity in the Universe, and so is the intermediary mass used to determine it.
While there are small masses in the universe like pebble sized rocks and debris in the asteroid
belt there are larger masses like planets. The intermediary mass is determined by a proton and
white dwarf star. It combined with the intermediary velocity 1/k which gives an energy that
predicts the life span of the universe in the standard Friedman model predicted by when the
last stars would be made and burn out from the mass of the Universe.!
This intermediary mass and energy seems to be associated with the masses and speeds in the
human realm between atoms and stars, like a mean for mass and velocity in the universe. It
predicts the radius and charge of a proton as well as the life span of the universe. But I asked
what do we know of that has exactly these masses and velocities. !
The intermediary mass is 68.897 kg is about 152 pounds. I knew a wolf is less than this a
jaguar more, but I found it is exactly an adult male mountain lion, a cougar.!
But what speed is 1/k=788.4626m/s. It is 1,764 miles per hour. That is the speed of a fast
military jet. it is about mach 2 or twice the speed of sound, which is about 767 miles per hour.
Our fastest jet (record holder) went mach 6.7. But it was a real milestone to achieve mach 2
which was first achieved by the Lockheed F-104 Starfighter.!
The fastest running animal is the Cheetah. So it would be like the record holding jet at mach
6.7. But the Cougar for which the adult male is nearly exactly the intermediary mass in the
Universe, would be like the F-104 Starfighter which goes almost exactly 1/k.!
We have suggested that the unit of a second and the unit of a meter are Natural and pertain to
the structure of the Universe. But what would be the Natural unit for mass. We could suggest it
is a kilogram, but that does not necessarily make sense. We have described the second in
terms of the proton!
!
And in terms of the Moon/Earth/Sun system!
!
And the meter in terms of the proton Earth/Moon/Sun system!
We want to describe the kilogram as a Natural unit. We already have the kilogram described as a
natural unit. It is described in terms of the density of water at standard temperature and
pressure:
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
K E
moon
K E
earth
(Ear th Day) = 1.08secon d s
(
M
⊙
M
e
⋅
r
e
r
m
⋅ r
p
r
e
)
1
2
KE
m
KE
e
(Ear th Day)
⋅
1
6α
2
r
p
m
p
h 4π
Gc
= 1.0m eters
of 37 61
But we want something more natural (Sure water is natural and a primary ingredient of life, and
standard temperature and pressure is ideal for life) but we want things in terms of the proton
and Earth/Moon/Sun system because we want to suggest the Earth/Moon/Sun system has a
Universal idea behind it and the proton is already Universal.
The mass of the Moon
The mass of the Earth
The intermediary mass
This needs to be multiplied by 1.2 to get 1 kg. But remember
It is actually closer to 1.25 seconds. And, remember
Thus our equation is
, ,
Using mean orbital velocities for the Earth and the Moon
1kg ≡ ρ
H2O
⋅ m eter
3
M
m
= 7.34767E 22kg
M
e
= 5.97219E 24kg
m
i
= 68.897kg
M
m
M
e
(m
i
) = (0.0123)(68.897kg) = 0.847kg
K E
m
K E
e
(Ear th Day) = 1.2secon d s
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 1.004996352secon d s
6α
2
⋅
M
m
M
e
(m
i
)
K E
m
K E
e
(Ear th Day) ⋅
m
p
r
p
Gc
h 4π
= 1kg
α = 1/137
1/α
2
= 1/18769
6/α
2
= 0.00032
K E
e
=
1
2
(5.97219E 24kg)(29,790m /s)
2
= 2.65E 33J
K E
m
=
1
2
(7.34767E 22kg)(2,033m /s)
2
= 3.837E 28J
K E
m
K E
e
(Ear th Day) =
3.837E 28
2.65E 33
(86,400s) = 1.251E 28secon d s
of 38 61
Remember using the sidereal day just makes it closer to 1.2s but makes little difference in the
results. We have
Thus for units of mass, length, and time as Natural Units we have…
!
!
And we point out that (because it really is quite incredible) for the unit of time the second is
described by just the proton, independently of the Earth/Moon/System and is described by just
the Earth/Moon/System, independently of the proton. In the case of the meter and the kilogram
they are described in terms of both, not independently of one another. The incredible thing
about the second is it comes from the evolution of the calendar since ancient times by
reconciling the periodicities of the Moon and Sun in the Earth sky by using sexagesimal, or base
60 counting, which we already went into in full. We have that
Then force, F, or is
(0.847)(1.251)
1
1.005s
= 1.054kg ≈ 1.0kg
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
K E
m
K E
e
(Ear th Day) = 1secon d
(
M
⊙
M
e
⋅
r
e
r
m
⋅ r
p
r
e
)
1
2
KE
m
KE
e
(Ear th Day)
⋅
1
6α
2
r
p
m
p
h 4π
Gc
= 1.0m
6α
2
⋅
M
m
M
e
(m
i
)
K E
m
K E
e
(Ear th Day) ⋅
m
p
r
p
Gc
h 4π
= 1kg
(
M
⊙
M
e
⋅
r
e
r
m
⋅ r
p
r
e
)
1
2
KE
m
KE
e
(Ear th Day)
= 1.0m /s
6α
2
(
M
⊙
M
e
⋅
r
e
r
m
⋅ r
p
r
e
)
1
2
KE
m
KE
e
(Ear th Day)
m
p
r
p
Gc
4πh
= 1.0m /s
2
kg
m
s
2
= 1N
of 39 61
We finally have energy, Joules, or is
Something very interesting is going on here so we want to start a new page to look at it so all the
equations are on one page…
36α
4
⋅
M
m
M
e
(m
i
)
M
⊙
M
e
⋅
r
e
r
m
⋅ r
p
r
e
1/2
m
2
p
r
2
p
Gc
4πh
= 1N
J = kg
m
2
s
2
6α
2
M
m
M
e
(m
i
)(m
p
)
Gc
4πh
⋅
M
⊙
r
e
M
e
r
m
⋅ r
e
KE
m
KE
e
(Ear th Day)
= 1J
of 40 61
Let us look at our equation for kilograms. This is actually incredible. Let’s see that: We write it
And say that
The mass above is the intermediary mass adjusted by the ratio of the Moon’s mass to the Earth’s
mass. The spin is given by the rotation of the Earth adjusted by the kinetic energy of the Moon to
the kinetic energy of the Earth and is close to one second, and in some cases is one second. The
frequency is given by the size and mass of a proton and is one cycle per second, its inverse is
actually one second. The product of all three is actually one kilogram even. The second turned
out to be a natural unit, and now so is the kilogram. We want to point out that the intermediary
mass is, where we made the approximation 0.77~3/4:
The constant k is
Which putting in is
And finally we see k with the velocity of the Earth defines the sixfold symmetry in this
(
M
m
M
e
(m
i
)
)(
K E
m
K E
e
(Ear th Day)
)(
6α
2
⋅
m
p
r
p
Gc
h 4π
)
= 1kg
(
M
m
M
e
(m
i
)
)
= m a ss = 0.847kg
(
K E
m
K E
e
(Ear th Day)
)
= spin = 1.2secon d s
(
6α
2
⋅
m
p
r
p
Gc
h 4π
)
= f requen c y = 1secon d
−1
m
i
m
i
=
3
2
c
3
h
3
8π
3
G
3
m
2
p
1/2
= 67.9943kg
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼 =
1
773.5
⋅
s
m
m
i
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
v
e
k v
e
= 6
m
i
(
1
k
)
2
= (68.897kg)(788.46m /s)
2
= 4.92558E6J ≈ 5,000,000J
of 41 61
In so-much-as we have
Where
And
Is like a quantum number where
And
Since quantum mechanical states are described by mass, spin, and frequency, it may be that the
Earth/Moon/Sun system is a quantum mechanical system. Which is interesting because the
Earth is the planet in our solar system brimming with life and the Moon makes that possible by
holding the Earth at the inclination to its orbit around the Sun that it has allowing for the
seasons preventing extreme hot and extreme cold. Thus we want to introduce the energy
And note that in a quantum mechanical system, the energy of a quantum is
Where is Planck’s constant and is frequency.
(
M
m
M
e
(m
i
)
)(
K E
m
K E
e
(Ear th Day)
)(
6α
2
⋅
m
p
r
p
Gc
h 4π
)
= 1kg
(
M
m
M
e
(m
i
)
)
= m a ss = 0.847kg
(
K E
m
K E
e
(Ear th Day)
)
= spin = 1.2secon d s
(
6α
2
⋅
m
p
r
p
Gc
h 4π
)
= f requen c y = 1secon d
−1
k v
e
= 6
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼 =
1
773.5
⋅
s
m
m
i
=
3
2
c
3
h
3
8π
3
G
3
m
2
p
1/2
= 67.9943kg
E = m
i
(
1
k
)
2
= (68.897kg)(788.46m /s)
2
= 4.92558E6J ≈ 5,000,000J
E = h ν
h
ν
of 42 61
We apply our formulation to predict the Earth orbital period as a quantum mechanical system.
In so far as we have for the solar system
And, for the the atom
We introduce what we will call a universal energy
The mass of the Moon is
We have
This is the Earth year (Earth orbital period) to an accuracy of
90%
E = m
i
(
1
k
)
2
= (68.897kg)(788.46m /s)
2
= 4.92558E6J ≈ 5,000,000J
E = h ν
h
G
⋅
c
3
m
p
=
6.62607E − 27
6.67408E − 11
⋅
(299792459)
3
1.67262E − 27
= 1.599E 36J ≈ 1.6E 36J
M
m
= 7.34767E 22kg
E = M
m
(
1
k
)
2
= 7.34767E 22kg(788.4626m /s)
2
= 4.56785E 28J
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
⋅
1
6α
2
⋅
r
p
m
p
4πh
Gc
=
1.599E 36J
4.56785E 28J
⋅ 1secon d = 35005527secon d s
(365.25d a ys)(24hrs /d a y)(60m /hr)(60sec /min) = 31557600secon d s
31557600
35005527
=
of 43 61
8.0 Solar Luminosity A QM State That I have found the Earth/Moon/Sun system is
quantum mechanical like the atom, is interesting because we find the orbital period of the Earth
around the Sun from the mass of the Moon orbiting it, alone. In classical mechanics, to get the
orbital period of the Earth you have to use the mass of the Sun and the Earth’s distance from it.
To explain this curious phenomenon that I have found regarding the Earth I am looking at why it
is the distance it is from the Sun because that is given by the orbital period (Kepler’s law,
). I am suggesting it is the distance it is from the Sun because it allows for life (puts it
in the habitable zone, where water exists in its liquid phase abundantly at a good temperature).
This is determined by the luminosity of the Sun. The luminosity of the Sun at its surface is
determined by its temperature at its core which is determined by its mass and radius. But the
temperature at the surface comes from the temperature at the core cooling by the time it
reaches the surface. We want to explain!
!
That is, the Earth year is given by the Moon’s mass because The Earth is ideal for life. The
temperature of the Sun at its core is described by!
1. !
= !
But we don’t want to just consider the mass of hydrogen ( ) but the mass of helium as well.
The sun is 91.0% hydrogen and 8.9% helium by number of atoms. We have!
0.91(1.67E-27)=1.5197E-27!
0.089(1.67E-27)4=5.9452E-28!
1.519E-27+5.9452E-29=2.11422E-27~2.11433E-27kg!
!
Heat dissipates by an inverse square law. The core of the Sun is 1.33E6m and its radius is
6.957E8m. Thus the temperature at its surface would be if the only factor was the inverse
square law (which it isn’t, it is complex and requires the inclusion of many factors):!
2. !
= !
T
2
= a
3
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
⋅
1
6α
2
⋅
r
p
m
p
4πh
Gc
= 1year
T
⊙
=
m
p
R
⊙
⋅
G
k
B
M
⊙
1.67E − 27kg
6.957E8m
⋅
6.67E − 11
1.38E − 23
1.989E30kg = 2.3E 7
∘
K
m
p
2E − 27kg
6.957E8m
⋅
6.67E − 11
1.38E − 23
1.989E30kg = 2.7637E 7
∘
K
T
S
=
(
R
c
R
s
)
2
T
⊙
(
1.22E6m
6.957E8m
)
2
(2.7637E 7 ∘ K )
of 44 61
= !
The actual temperature at the surface of the Sun is 5,780 degrees K. So we are off by a factor
of!
!
It is at this point that we suggest we can bypass the complex factors that also affect the
surface temperature of the Sun by using our Quantum Mechanical system for the solar system.
Not that the Sun isn’t described by Quantum mechanics in looking at its atoms, but in that our
Quantum Mechanical system treats the sun as if it is one proton with planets orbiting it. To do
this we notice that the factor of 68 is very close to our intermediary mass . But that is
in kilograms and our factor is dimensionless. But we developed a quantum mechanical
description of 1kg in the last section for the Earth/Moon/Sun system. It was!
3. !
Equation 1, 2, and 3 yield!
4. !
=
!
=(0.000003075)(2.7637E7deg K)((81.28)(0.804)=5553.6 deg K!
The luminosity of the Sun is given by!
5. !
= !
The actual luminosity of the Sun is 3.846E26 Watts. This was using the second given by Earth
and Moon kinetic energies as 1.2 seconds from average velocities. We can go as low as 1.08
seconds using apogees and perigees. And we can get values in between because though the
orbits of the Moon and Earth are close to circular, they aren’t exactly, so there is slight
variations in the result, so the above wavers around the exact answer.!
We want to consider how much sunlight reaches the Earth. We use our value for solar
luminosity:!
!
84.98975 ≈ 85
∘
K
5780
85
= 68
68.897kg
(
M
m
M
e
(m
i
)
)(
KE
m
KE
e
(Ear th Day)
)(
6α
2
⋅
m
p
r
p
Gc
h4π
)
= 1kg
T
S
=
(
R
c
R
s
)
2
(
m
p
R
⊙
⋅
G
k
B
M
⊙
)
1
6α
2
⋅
r
p
m
p
4πh
Gc
M
m
M
e
⋅
KE
m
KE
e
(Ear th Day)
(
1.22E6m
6.957E8m
)
2
(
2E − 27kg
6.957E8m
⋅
6.67E − 11
1.38E − 23
(1.989E30kg)
)(
5.97219E 24kg
7.34767E 22kg
)
1.005s
1.25s
L
⊙
= 4πR
2
⊙
σ T
4
S
4π(6.957E8m)
2
(5.67E − 8)(5553.6
∘
K )
4
= 3.28E 26Wat ts
L
0
= 3.28E 26J/s
of 45 61
The separation between the Earth and the Sun is!
!
The solar luminosity is reduced at Earth by the inverse square law giving the solar constant:!
!
intercepts the Earth disc and distributes itself over the entire surface and
because the Earth’s albedo, , is 0.3 it reflects 30% of the light back into space. We have!
!
Temperature to the fourth is proportional to radiation by the Steffan-Boltzmann constant, :
means temperature entering:!
!
So we have the annual average temperature would be!
!
!
!
The solar constant is usually 1361 (Well established). Let’s convert this to centigrade:!
!
!
The actual annual average temperature of the Earth is 14C or 57F. But we have not included
the effect of the Earth’s greenhouse gases holding in heat. In a one layer atmosphere model we
consider that the radiation entering the system equals the radiation leaving the system when
the Earth is in radiative equilibrium. The atmosphere radiates back to the surface. So
the temperature at the surface of the Earth is!
!
r
e
= 1.496E11m
S
0
=
3.28E 26
4π(1.496E11m)
2
= 1166
Wat ts
meter
2
S
0
πR
2
4πR
2
a
(1 − a)S
0
πR
2
4πR
2
= (1 − a)
S
0
4
σ
T
e
σ T
4
e
=
S
0
4
(1 − a)
T
e
=
4
S
0
4σ
(1 − a)
σ = 5.67E − 8
T
e
=
4
1166
4(5.67E − 8)
(0.7) = 245K
T
e
= 245K − 273.15K = − 28.15
∘
C
(−28.15)(1.8) + 32 = − 18.67
∘
F
T
a
= T
e
σ T
4
S
= σ T
4
e
+ σ T
4
a
= 2σ T
4
e
of 46 61
So,…!
!
!
!
!
So this is hotter than the annual average temperature we said was 14C or 57F but we have not
considered cooling by convection; A lot of the radiation goes into warming the ocean, which
has a high specific heat (4.184J/g-K) and when this water evaporates it precipitates and returns
as rain. If we include this we get much closer to accurately predicting the global temperature.
There are other cooling mechanisms to consider as well. !
We have done what we set out to do to show it is possible that the Earth year is given by the
mass of the Moon because the Earth/Moon/Sun system is like a quantum mechanical one that
provides a state optimum for life. We have shown!
!
Where , hydrogen, is weighted with , helium according to their relative abundances by
number of atoms of each.
T
S
= 2
1
4
T
e
T
S
= (1.18921)245 = 291.356K
T
S
= 302.75 − 273.15 = 18.206
∘
C
18.206(1.8) + 32 = 64.771
∘
F
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
⋅
1
6α
2
⋅
r
p
m
p
4πh
Gc
= 1year
T
S
=
(
R
c
R
s
)
2
(
m
p
R
⊙
⋅
G
k
B
M
⊙
)
1
6α
2
⋅
r
p
m
p
4πh
Gc
M
m
M
e
⋅
KE
m
KE
e
(Ear th Day)
m
p
4m
p
of 47 61
9.0 Genesis Project Equations We have for the Earth/Moon/Sun system for there to be life on
Earth that the temperature of the Sun at its surface is:!
!
And that the Earth orbital period is:!
We let the following ratio be approximated as 1:
So we have
!
We notice that . That is, the radius of the core of the Sun is on the order of the radius
of the Moon. the radius at the surface of the Sun, is the solar radius . We find
must be multiplied by a factor somewhere between the golden ratio and 2/3 the fibonacci
ratio that approximates it. We find that value is . Thus our equation for the
temperature of the Sun is:!
!
This gives us!
!
From this we can compute the luminosity of the Sun:!
!
T
S
=
(
R
c
R
S
)
2
(
m
p
R
⊙
⋅
G
k
B
M
⊙
)
1
6α
2
⋅
r
p
m
p
4πh
Gc
M
m
M
e
⋅
KE
m
KE
e
(Ear th Day)
T
earth
=
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
⋅
1
6α
2
⋅
r
p
m
p
4πh
Gc
1
6α
2
⋅
r
p
m
p
4πh
Gc
KE
m
KE
e
(Ear th Day)
≈ 1
T
S
=
(
R
c
R
S
)
2
(
m
p
R
⊙
⋅
G
k
B
M
⊙
)
M
e
M
m
R
c
≈ Rm
R
S
R
⊙
R
m
/R
⊙
ϕ
4
2
/5
2
= 16/25
T
S
=
(
R
m
R
⊙
)
2
(
m
p
⋅
G
k
B
⋅
M
⊙
R
⊙
)
M
e
M
m
T
S
=
(
16
25
⋅
1.7374E6
6.957E8
)
2
(
2E − 27
6.67E − 11
1.38E − 23
⋅
1.989E30
6.957E8
)(
5.972E 24
7.34767E 22
)
= 5,738
∘
K
L
⊙
= 4πR
⊙
σ T
4
S
of 48 61
= !
The actual solar luminosity is 3.846E-26 watts, so this is a very good estimate. We have our
system of equations for a Earth/Moon/Sun life bearing system is!
!
We can call these the Genesis Project equations after the Genesis project in the Star Trek movie
where they created a technology that converted dead worlds into living worlds."
4π(6.957E8)
2
(5.67E − 8)(5,738
∘
K )
4
= 3.738E 26wat ts
L
⊙
= 4πR
2
⊙
σ
(
R
m
R
⊙
)
2
(
m
p
⋅
G
k
B
⋅
M
⊙
R
⊙
)
M
e
M
m
4
T
earth
=
h
G
⋅
c
3
m
p
M
m
(
1
k
)
2
⋅
1
6α
2
⋅
r
p
m
p
4πh
Gc
S
0
=
L
⊙
4π r
2
e
T
2
earth
= r
3
e
m
i
=
3
2
c
3
h
3
8π
3
G
3
m
2
p
1/2
= 67.9943kg
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼 =
1
773.5
⋅
s
m
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
k v
e
= 6
of 49 61
10.0 Another Theory For The Radius Of A Proton
The ancient Greeks knew motion to be an illusion. We see this in Zeno’s arrow. An arrow is fired
so as to travel from A to B. Since to go to B it must travel to (1/2)B, then to half way to B from
there or to 1/4, then to halfway from 1/4 to B or to 1/8, and so on. Since we can keep dividing
the successive distances distances in half that must be traveled, the arrow never gets to B.
Mathematically this is!
!
!
!
!
!
!
As the number of partitions, , go to infinity their widths go to zero and distances converges on
one. But for Zeno’s arrow the number of partitions never go to infinity so their widths never go
to zero, so the distance never converges on one and we see that motion is an illusion. The
conundrum is where we have said . Is two raised to the zero really one? It was
established as a convention, there is no explanation for it. Couldn’t it just as easily be zero (two
times itself zero times is 0)? Or two times itself zero times is two? In!
n
∑
i=1
i = 1 + 2 + 3 + . . . + n =
n(n + 1)
2
lim
n⇀∞
= ∞
n
∑
i=1
(
1
2
)
i
=
(
1
2
)
1
+
(
1
2
)
2
+
(
1
2
)
3
+ . . . =
1
2
+
1
4
+
1
8
+ . . .
lim
n⇀∞
n
∑
i=1
(
1
2
)
i
=
(
1
2
)
1
+
(
1
2
)
2
+
(
1
2
)
3
+ . . . =
1
2
+
1
4
+
1
8
+ . . . = 1
lim
n⇀∞
n−1
∑
i=0
(
i
n
)
=
n(n + 1)
n
n−1
∑
i=0
(
1
2
i
)
= 2 −
1
2
n−1
lim
n⇀1
n−1
∑
i=0
(
1
2
i
)
= 2 −
1
2
1−1
= 2 −
1
2
0
= 1
n
2
0
= 1
of 50 61
we can say n can always take on a larger value so there can always be another
partition. If were zero then and any distance is infinite, can never be traversed.!
In traveling from A to B something in motion at some time passes through all points but never
exists at any of them for any amount of time because there and infinite number of points
between A and B and they have all been passed through in a finite amount of time. This is only
possible if the object in motion wasn’t at any point for any amount of time, or the journey would
have taken forever. Motion is adding up an infinite amount of zeros adding up to a finite time.
Velocity is then the rate at which something passes through a point. Since an object in motion
exists at a point for no time regardless of its velocity, the time it is at one point is the same if its
velocity was different. Something only exists at a given point for an amount of time if it has
stopped moving. So velocity must be described by energy, the energy it has at a point at a
given time. Thus we have Planck’s constant is or energy over time. Thus
energy is described by frequency:!
, , , , !
Then the rest frequency of a proton is !
10.1. !
10.2. !
We have!
10.3. !
If we take the radius of a proton to be!
0.8438E-15m which is very close to the CODATA results around 2018 (0.842E-15m) the we
have!
10.4. !
10.5. !
The result is that the radius of a proton is, given its mass!
10.6. !
10.7. , 10.8. !
n(n + 1)
2
2
0
1
0
= ∞
h
Joule ⋅ secon d s
E = h f
f = 1/s
h = J ⋅ s
h f = (J ⋅ s)(1/s)
E = J = Joules = energy
E = (1.67262E − 27kg)(299792459m /s)
2
= 1.503275897E − 10J
f =
E
h
=
1.503275897E − 10J
6.62607E − 34J ⋅ s
= 2.3687E 23s
−1
f
p
=
m
p
c
2
h
0.8438E − 15m
299,792,459m /s
= 2.8146E − 24s
(2.3687E 23s
−1
)(2.8146E − 24s) ≈ 0.6667 = 2/3
m
p
r
p
=
2
3
⋅
h
c
r
p
=
2
3
⋅
h
cm
p
m
p
=
2
3
⋅
h
cr
p
of 51 61
The radius of a proton is not well known, but are all well known.!
We found using aphelions and perihelions:!
Equation 1.6.
And we had equation 1.4
1.4.
Using 10.7 and 10.8
10.7. , 10.8.
And equation 1.4, we can describe the kinetic energies of the Moon and Earth with respect to
Earth rotation (Earth Day) in terms of proton mass and in terms of proton radius, as well as
come up for expressions for one second in terms of each of these.
10.9
10.10.
10.11.
10.12.
And dividing one by the other, we have
10.13.
That is
h, c, m
p
K E
moon
K E
earth
(Ear th Day) = 1.08secon d s
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
r
p
=
2
3
⋅
h
cm
p
m
p
=
2
3
⋅
h
cr
p
1
9
⋅
1
α
2
m
2
p
⋅
h
c
h 4π
Gc
= 1.062836secon d s
1
4α
2
⋅
r
p
c
h
h 4π
Gc
= 0.94536secon d s
1
9
⋅
1
α
2
m
2
p
⋅
h
c
h 4π
Gc
=
K E
m
K E
e
(Ear th Day)
1
4α
2
⋅
r
p
c
h
h 4π
Gc
=
K E
m
K E
e
(Ear th Day)
4
9
⋅
h
2
r
2
p
m
2
p
⋅
1
c
2
≈ 1
1.062836secon d s
0.94536secon d s
= 1.124
of 52 61
Which is about 89% accuracy. At this point we should write out what we had from the outset of
this paper combining equation 1.4 and 1.6!
Equation 10.14. !
And now we have!
Equation 10.11.
Equation 10.12.
And,…
Equation 1.4.
Equation 1.6.
And we now have!
10.9
10.10.
We see again that our Moon orbiting our Earth is pivotal to the Universe, and that the second is
a natural unit of measurement. If
10.13.
Is supposed to be exactly 1, then the radius of a proton is
10.15.
"
1
6α
2
r
p
m
p
h4π
Gc
=
KE
m
KE
e
(Ear th Day)
1
9
⋅
1
α
2
m
2
p
⋅
h
c
h 4π
Gc
=
K E
m
K E
e
(Ear th Day)
1
4α
2
⋅
r
p
c
h
h 4π
Gc
=
K E
m
K E
e
(Ear th Day)
1
6α
2
r
p
m
p
h 4π
Gc
= 1.0044996352
K E
moon
K E
earth
(Ear th Day) = 1.08secon d s
1
9
⋅
1
α
2
m
2
p
⋅
h
c
h 4π
Gc
= 1.062836secon d s
1
4α
2
⋅
r
p
c
h
h 4π
Gc
= 0.94536secon d s
4
9
h
2
r
2
p
m
2
p
⋅
1
c
2
≈ 1
r
p
=
2
3
⋅
h
m
p
⋅
1
c
r
p
=
2
3
⋅
(6.62607E − 34)
(1.67262E − 27)
⋅
1
299,792,459
= 0.880941E − 15m
of 53 61
11.0 Macro and Micro Planck Constants!
We found using aphelions and perihelions:!
Equation 1.6.
And we had equation 1.4
1.4. !
We write 1.6 in a form of Joule-seconds so we have a Planck constant given by the Earth orbit
which is determined by the Sun, we will call it the solar Planck constant which we will
distinguish from Planck’s constant by making the Planck constant the proton Planck
constant, . We have!
11. 1. !
11.2. !
From 1.4 we have!
11.3. !
From 10.6 we have!
10.6. !
11.4. !
These give!
11.5. !
From 11.4 we have!
11.6. !
K E
moon
K E
earth
(Ear th Day) = 1.08secon d s
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
h
⊙
h
h
p
KE
earth
(1secon d ) = h
⊙
= J ⋅ s
1sec =
h
⊙
KE
e
h
p
= (1sec)
2
(36α
4
m
p
)
Gc
4πr
2
p
m
p
r
p
=
2
3
⋅
h
c
h
p
=
3
2
r
p
m
p
c
h
⊙
=
KE
e
α
2
1
6
⋅
r
3
p
c
m
p
⋅
π
Gc
h
⊙
h
p
=
2
3
⋅
KE
e
c
⋅
1
α
2
1
6
⋅
r
p
m
3
p
⋅
π
G
of 54 61
!
= !
Using aphelions and perihelions!
Multiply that by our 1 second as a natural unit.!
!
If we use!
!
!
!
We can write equation 11.6 as!
11.7 !
This gives us that is given by some constant times the kinetic energy of the Earth. Let
us see what that constant is:!
11.8. !
= !
= This gives!
!
h
⊙
= (6.62607E − 34)
2
3
2.7396E33
299,792,459
⋅ 18769
1
6
⋅
(0.833E − 15)
1.67262E − 27)
3
⋅
π
(6.67408E − 11)
2.8314E33J ⋅ s
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
h
⊙
=
2.7396E33J
2.8314E33J
100 = 96.76 %
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
h
⊙
= (2.7396E 33J )(1.004966s) = 2.7533E 33J ⋅ s
h
⊙
=
2.7533E33J
2.8314E33J
100 = 97.24 %
h
⊙
h
p
=
1
3
⋅
KE
e
α
2
c
1
3
⋅
2π r
p
Gm
3
p
h
⊙
/h
p
C
C =
1
3
⋅
1
α
2
c
1
3
⋅
2π r
p
Gm
3
p
1
3
⋅
18769
299792459
1
3
2π (0.833E − 15
(6.67408E − 11)(1.67262E − 27)
3
1.5597656E33J
−1
≈ 1.56E 33J
−1
h
⊙
= h
p
(1.56E33J
−1
)KE
e
= (6.62607E − 34)(1.56E33)(2.7396E33)
of 55 61
We notice that !
So we have!
11.9. !
Where!
11.10. !
Which means is an Eigenvalue because it maps something into itself because !
11.12. !
Because!
!
Is approximately 1 second. In quantum mechanics eigenvalues describe atomic energy states,
discreet orbits of electrons around a proton. Here we see that is something happening but with
the Earth orbiting the Sun to relate the solar Planck constant on the macro-scale to the proton
Planck constant on the micro-scale. We can also say that!
11.13. !
Which makes an eigenvalue as well because it maps into because is the
macro-scale analog of . That is the incredible thing about !
!
Being approximately 1 second where 1 is the identity mapping something into itself. Thus we
have shown that the Earth/Sun/Moon orbital system is a quantum mechanical system
connected to the the proton/electron quantum mechanical systems. We include the Moon
because!
Is approximately 1 second. We have shown what we set out to do as well, and that is show that
the second is a natural unit, and a natural constant. We found it is
h
p
C = 1.033667s ≈ 1secon d
h
⊙
= (h
p
C )KE
e
(h
p
C ) = 1secon d
h
p
C
h
⊙
= K E
e
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
h
⊙
= (C ⋅ K E
e
)h
p
C ⋅ KE
e
h
⊙
h
p
h
⊙
h
p
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
K E
moon
K E
earth
(Ear th Day) = 1.08secon d s
of 56 61
11.14.
11.15. !
1secon d = h
p
C
C =
1
3
⋅
1
α
2
c
1
3
⋅
2π r
p
Gm
3
p
of 57 61
12.0 The Exact Radius of a Proton We may be able to come up with the actual value of the
radius of a proton by suggesting space and time are such that it is given by the golden ratio, .
Since we have 1 second is given by equation 1.4
Eq 1.4.
And we now have that one second is given by as well
Eq 12.1.
Then we have
Which is
Equation 12.2.
Which is our equation 10.15 for the radius of a proton
10.15.
We see this gives
Is an accuracy of 94.56%. We suggest that the nature of space and time is such that the equation
holds if is supposed to be . Then since , , and are well known, we
would be able to determine exactly what the the radius of a proton , is. Since
, we have
Equation 12.3.
This gives
Equation 12.4.
We see this is the momentum times distance of Quantum Mechanics. That is
Φ
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
1
3
⋅
h
α
2
c
1
3
⋅
2π r
p
Gm
3
p
= 1.06secon d s
(
1
6α
2
⋅ r
p
m
p
4πh
Gc
)
3α
2
1
⋅
c
h
3Gm
3
p
2π r
p
= 1
3
2
⋅
c
h
⋅ m
p
r
p
= 1
r
p
=
2
3
⋅
h
m
p
⋅
1
c
3
2
⋅
299,792,459
6.62607E − 34
⋅
1.67262E − 27)(0.833E − 15)
1
= 0.945579897 = 1
3/2
Φ = ( 5 + 1)/2
h
c
m
p
r
p
Φ = 1/ϕ = ( 5 − 1)/2 = 0.618034
ϕ
h
c
⋅
1
m
p
= r
p
r
p
= 0.816677E − 15m ≈ 0.817m
of 58 61
Equation 12.5.
Is
Equation 12.6.
Equation 12.7.
We find it in the Schrödinger wave equation
Equation 12.8.
But we see as well this the the asymptote of set , the clock hyperbola, in Special Relativity in
spacetime diagrams, it is the world-line of a light signal. This is the hyperbola of special
relativity given by
Equation 12.9.
See following illustrations….
cm
p
r
p
ϕh
= 1
cm
p
r
p
= (mom ent u m)(dista nce) = px
ϕh = (m om ent u m)(dista n ce) = px
(iℏ)∂Ψ = − p∂xΨ
g
t
2
−
(
x
c
)
2
= 1
of 59 61
of 60 61
of 61 61
The Author!